Question

Let $G$ and $H$ be affine algebraic groups over a scheme $S$ of characteristic 0 and let $\textbf{Hom}_{S,gp}(G,H)$ be the functor $T \mapsto \text{Hom}_{T,gp}(G,H)$

Theorem (SGA 3, expose XXIV, 7.3.1(a)): Suppose G is reductive. Then $\textbf{Hom}_{S,gp}(G,H)$ is representable by a scheme.

Can this fail if $G$ is not reductive? I worked out a few example with $G = \mathbb{G}_a$, but they were representable.

Answer 1

Hom(G_a, G_m) is not representable.

Let R be a ring of characteristic zero. I claim that Hom(G_a, G_m)(Spec R) is {Nilpotent elements of R}. Intuitively, all homs are of the form x -> e^{nx} with n nilpotent.

More precisely, the schemes underlying G_a and G_m are Spec R[x] and Spec R[y, y^{-1}] respectively. Any hom of schemes is of the form y -> \sum f_i x^i for some f_i in R. The condition that this be a hom of groups says that \sum f_k (x_1+x_2)^k = (\sum f_i x_1^i) (\sum f_j x_2^j). Expanding this, f_{i+j}/(i+j)! = f_i/i! f_j/j!. So every hom is of the form f_i = n^i/i!, and n must by nilpotent so that the sum will be finite.

Now, let's see that this isn't representable. For any positive integer k, let R_k = C[t]/t^k. The map x -> e^{tx} is in Hom(G_a, G_m)(Spec R_k) for every k. However, if R is the inverse limit of the R_k, there is no corresponding map in Hom(G_a, G_m)(Spec R). So the functor is not representable.

Answer 2

There is a reasonable salvage, at least if the base scheme is a field: Hom(G,H) is a direct limit of representable subfunctors. See Lemma A.8.13 in the book "pseudo-reductive groups" (where it is used to prove that the scheme-theoretic fixed locus for a linearly reductive group acting on a connected reductive group is always reductive (possibly disconnected) provided the base is a field.